B A light beam that never interacts with matter

I have read that infinite time dilation goes hand in hand with infinite length contraction for photons. AKA "photons do not experience time", or from a photons perspective, it takes no time to arrive at its destination.

What if you were floating in space with a laser pointer, and you pointed the laser somewhere in the direction of the intergalactic medium, held the button for 1 second, and let go. The probability of that photon beam hitting any matter would be pretty much zero. (?)

So if the photon does not have a destination (assuming it didn't interact with any matter or em waves), how can it reach its destination in no time?

The quick summary is that the concept of "experienced time" (the more precise term in relativity is "proper time") is meaningless for a photon; it simply doesn't apply. So you can't reason as though it does, which means you can't use "it takes no time to arrive" as a premise in your reasoning.

What if you were floating in space with a laser pointer, and you pointed the laser somewhere in the direction of the intergalactic medium, held the button for 1 second, and let go. The probability of that photon beam hitting any matter would be pretty much zero. (?)

"Pretty much" is not the same as "exactly". Any given piece of matter will subtend some finite solid angle, so the photon will have some small but finite chance of hitting it. But even if it doesn't, so what? It will just continue to propagate through the universe.

Someone wrote in another thread that a photon "is simply a quantized interaction between the EM wave and matter or another EM wave."

My question is, if the photon does not have anything to interact with like in the OP, and it is not acceptable to think of a photon as having the shape of an EM wave propagating through space...then what is it?

Staff: Mentor

My question is, if the photon does not have anything to interact with like in the OP, and it is not acceptable to think of a photon as having the shape of an EM wave propagating through space...then what is it?

If you aren't up for digging through that much math, an easy answer is to say that light is electromagnetic waves and photons only come into the picture during the interaction between these waves and matter. It turns out that the electromagnetic waves always deliver their energy in discrete amounts at a single point; and when that happens we say "a photon landed there".

Staff: Mentor

Aka what if an electron was located on the plane of the photons electric field, it would be accelerated according to F=qE?

Eleven minutes might not have been long enough to look carefully at those two references....

Yes, an electron in an electrical field is accelerated according to ##\vec{F}=-q\vec{E}##. However, it makes very little sense to talk of "the photon's electric field" in this situation. We have an electromagnetic wave, and that's all we need to calculate the interaction between it and the electron. Photons only come into the picture if we want a complete quantum electrodynamics treatment of the interaction, and that can't be done in English; it needs far more math than fits in a B-level thread. (And we'd end up with the same answer anyway - quantum electrodynamics reduces to classical electrodynamics whenever the quantum effects are small enough to ignore).

Eleven minutes might not have been long enough to look carefully at those two references....

Yes, an electron in an electrical field is accelerated according to ##\vec{F}=-q\vec{E}##. However, it makes very little sense to talk of "the photon's electric field" in this situation. We have an electromagnetic wave, and that's all we need to calculate the interaction between it and the electron. Photons only come into the picture if we want a complete quantum electrodynamics treatment of the interaction, and that can't be done in English; it needs far more math than fits in a B-level thread. (And we'd end up with the same answer anyway - quantum electrodynamics reduces to classical electrodynamics whenever the quantum effects are small enough to ignore).

I'll look at it, sorry I am just about to sleep.

I get that this is an oversimplification, but it seems like a contradiction to me. I'm sure it will come out when I read the reference. I just don't get how we can say that an electron is being continuously accelerated by the F=qE force if it is experiencing the field of say a radio wave, but the radio wave itself cannot continuously give its energy to the electron, it can only give energy quanta.

I have read that infinite time dilation goes hand in hand with infinite length contraction for photons.

Do you remember where? Because there is no such thing as infinite time dilation and infinite length contraction.

Time dilation and length contraction are phenomena that describe what happens when measurements taken using two different frames of reference are compared. Since reference frames never move at the speed of light, it doesn't make any sense to refer to measurements that might be taken in a frame of reference that moves at the speed of light.

Here is the reasoning used to arrive at a statement that comes as close as I can to yours.

The factor ##\gamma## describes the amount of time dilation and length contraction. $$\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}.$$ So what happens is that as the relative speed ##v## of two reference frames approaches the speed ##c## of light, ##\gamma## approaches infinity. Thus so does the amount of time dilation and length contration. But infinity is something approached, not reached. And ##v## can approach, but never reach, ##c##.

People naturally ask what would it would be like if ##v## actually reached ##c## and time dilation and length contraction became infinite. And of course that question cannot be answered because it's based on a false premise. ##v## cannot ever reach ##c##. So when we realize that light itself travels at speed ##c## we think that time dilation and length contraction are infinite in that case. But that is not a case, because light is not a reference frame.

And of course a mathematician would remind us that ##\gamma## is not infinite when ##v## equals ##c##. It's undefined. It increases beyond all bounds as ##v## approaches ##c## and we can describe that as "approaching infinity" but that is not a reason for ever claiming that it reaches infinity. The very notion of reaching infinity is absurd.

I did read through it. I don't really understand Hamiltonian mechanics, but what I take from it is that light is an infinite set of quantum harmonic oscillators. Somehow at the end he gets that the energy states are eigenvectors of the momentum and that's why they are quantized.

Nugatory's link said:

The states we are describing have a particle interpretation, but the theory is richer than just a collection of particles since, e.g., one can superimpose the states described above to get states which encode the field like properties of the EM field. Of course, such states will not be stationary states. Also, recall our discussion of the “fluctuations” of E and B in the vacuum. Charges respond to the electromagnetic field strengths (think: Lorentz force law), and so the behavior of charged particles can be affected by EM phenomena, even when no photons are present!

So I guess I'm still confused conceptually as to how an electron could be continuously accelerated by an EM wave, even though the EM wave can only change it's energy state in discrete steps. Does that mean that the electron is only accelerated in discrete steps?

Staff: Mentor

I'm still confused conceptually as to how an electron could be continuously accelerated by an EM wave, even though the EM wave can only change it's energy state in discrete steps.

This is not a confusion that can be resolved in a "B" level thread. The answers you have already been given are the best that can be done at the "B" level. Once you have spent some time looking at the references you have been given and have improved your understanding to the point where you can have an "I" level discussion, you can start a new thread at that level if necessary (but it might not be necessary).